Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}cos4x+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow-\dfrac{3}{4}-\dfrac{1}{4}\left(1-2sin^22x\right)+\dfrac{1}{2}sin2x=0\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b/ ĐKXĐ: \(cos2x\ne0\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(6sinx-2cos^3x=\frac{10sin2x.cos2x.sinx}{2cos2x}\)
\(\Leftrightarrow6sinx-2cos^3x=5sin2x.sinx\)
\(\Leftrightarrow3sinx-cos^3x=5cosx.sin^2x\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(3tanx\left(1+tan^2x\right)-1=5tan^2x\)
\(\Leftrightarrow3tan^3x-5tan^2x+3tanx-1=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(3tan^2x-2tanx+1\right)=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\) (ko thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(cos^3x+sin^3x=sin2x+sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(1-\dfrac{sin2x}{2}\right)=sin2x+sinx+cosx\\ \Leftrightarrow-\dfrac{1}{2}sin2x\left(sinx+cosx+2\right)=0\\ \)
Mà \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)>-2\)
\(\Rightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\)
Tổng các nghiệm của phương trình trong \(\left[0;2018\pi\right]\) là:
\(S=\dfrac{\left(0+2018\pi\right)\left(\dfrac{2018\pi-0}{\dfrac{\pi}{2}}+1\right)}{2}=4073333\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)
\(\Leftrightarrow\left(2cosx-2sinx\right)\left(sinx+cosx\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(\text{vô nghiệm trên đoạn xét}\right)\\2cosx-2sinx+sinx.cosx=m\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(t=cosx-sinx=\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}t\in\left[-1;1\right]\\sinx.cosx=\dfrac{1-t^2}{2}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2t+\dfrac{1-t^2}{2}=m\)
Xét hàm \(f\left(t\right)=-\dfrac{1}{2}t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)
\(-\dfrac{b}{2a}=2\notin\left[-1;1\right]\) ; \(f\left(-1\right)=-2\) ; \(f\left(1\right)=2\)
\(\Rightarrow-2\le f\left(t\right)\le2\Rightarrow-2\le m\le2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
ĐK : sinx cosx > 0
Khi đó phương trình trở thành
sinx+cosx=\(2\sqrt{\sin x\cos x}\)
ĐK sinx + cosx >0 → sinx>0 ; cosx>0
Khi đó \(2\sqrt{\sin x\cos x}\Leftrightarrow2\sin x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
Vậy ...
Bài 2:
ĐK : \(\sin\left(3x+\frac{\pi}{4}\right)\ge0\)
Khi đó phương trình đã cho tương đương với phương trình \(\sin2x=\frac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
Trong khoảng từ \(\left(-\pi,\pi\right)\) ta nhận được các giá trị :
\(x=\frac{\pi}{12}\) (TMĐK)
\(x=-\frac{11\pi}{12}\) (KTMĐK)
\(x=\frac{5\pi}{12}\) (KTMĐK)
\(x=-\frac{7\pi}{12}\) (TMĐK)
Vậy ta có 2 nghiệm thõa mãn \(x=\frac{\pi}{12}\) và \(x=-\frac{7\pi}{12}\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=4\left(sinx+cosx\right)\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-4\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(-3-\frac{1}{2}sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\sin2x=-6\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow cosx=-sinx=cos\left(\frac{\pi}{2}+x\right)\)
\(\Rightarrow x=-\frac{\pi}{2}-x+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
\(0\le-\frac{\pi}{4}+k\pi\le\pi\Rightarrow k=1\)
\(\Rightarrow x=\frac{3\pi}{4}\)